cho biểu thức M = \(\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) . Khi x > 0 ; x≠1.
Rút gon biểu thức M
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26 tháng 10 2021
a) \(M=\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(x+1\right)}.\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) \(x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}+2-\sqrt{3}=4\)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{4}+1}{\sqrt{4}-1}=\dfrac{2+1}{2-1}=3\)
Y
25 tháng 7 2023
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
8 tháng 7 2021
\(M=\left(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) với x>0;x≠1
\(=\left(\dfrac{x\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}\)
\(M=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{x}=\dfrac{-x+\sqrt{x}+2}{x\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)}{x\left(\sqrt{x}+1\right)}=\dfrac{2-\sqrt{x}}{x}\)
vậy M=\(\dfrac{2-\sqrt{x}}{x}\)
vì x>0 nên để \(M< 0\Leftrightarrow\dfrac{2-\sqrt{x}}{x}< 0\Leftrightarrow2-\sqrt{x}< 0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)
6 tháng 7 2021
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\text{x > 0, x ≠ 1}\)
\(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{x-1-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
12 tháng 4 2022
1.\(x=4\)
\(B=\left(\dfrac{x+1}{2}-\sqrt{x}\right)=\left(\dfrac{4+1}{2}-\sqrt{4}\right)=\dfrac{5}{2}--2=\dfrac{5-4}{2}=\dfrac{1}{2}\)
2.\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)=\left(\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+1}{2}-\sqrt{x}=\dfrac{x+1-2\sqrt{x}}{2}=\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(M=A.B=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
3.\(M=\dfrac{\sqrt{x}}{6}\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{6}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=6\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+\sqrt{x}=6\sqrt{x}-6\)
\(\Leftrightarrow x-5\sqrt{x}+6=0\)
Đặt \(\sqrt{x}=a;a\ge0\)
=> pt trở thành:
\(a^2-5a+6=0\)
\(\Delta=\left(-5\right)^2-4.6=25=24=1>0\)
=> pt có 2 nghiệm:
\(\left\{{}\begin{matrix}x_1=\dfrac{5+\sqrt{1}}{2}=3\left(tm\right)\\x_2=\dfrac{5-\sqrt{1}}{2}=2\left(tm\right)\end{matrix}\right.\)
Xét \(\sqrt{a}=3\)
\(\Leftrightarrow a=9\)
Xét \(\sqrt{a}=2\)
\(\Leftrightarrow a=4\)
Vậy \(x=9;4\)
7 tháng 7 2021
\(=>A=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left[\dfrac{\sqrt{x}+1-2}{x-1}\right]\)
\(=>A=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}\)
b,\(=>\dfrac{1}{\sqrt{x}}=\dfrac{1}{2}=>\sqrt{x}=2=>x=\sqrt{2}\left(tm\right)\)
16 tháng 11 2021
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
AH
Akai Haruma
Giáo viên
26 tháng 5 2023
Lời giải:
\(P=\left[\frac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}+\frac{x}{\sqrt{x}(\sqrt{x}-1)}\right]:\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left[\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}-1}\right].\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}.\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{\sqrt{x}}{\sqrt{x}-1}\)
b. Áp dụng BĐT AM-GM
\(M=P\sqrt{x}=\frac{x}{\sqrt{x}-1}=\frac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}-1}\)
\(=(\sqrt{x}-1)+\frac{1}{\sqrt{x}-1}+2\geq 2\sqrt{(\sqrt{x}-1).\frac{1}{\sqrt{x}-1}}+2=2+2=4\)
Vậy $M_{\min}=4$ khi $\sqrt{x}-1=\frac{1}{\sqrt{x}-1}$
$\Rightarrow \sqrt{x}-1=0$
$\Leftrightarrow x=1$
\(M=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}\cdot\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ M=\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ M=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}+1}\)
\(M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x-1-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}\right)}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)